of the following, ____ has the lowest boiling point
cl2
o2
n2
h2

The root-mean square speed of the CO₂ molecules is 339 m/s.

The given parameters;

The molecular mass of the CO₂ = (12 + 16x2) = 44 g = 0.044 kg

The average mass of the CO₂ molecule is calculated as;

[tex]average \ mass \ = \frac{molecular \ mass}{Avogadro's \ number} = \frac{0.044}{6.02 \times 10^{23}} = 7.31 \times 10^{-26} \ kg[/tex]

The root-mean square speed is calculated by applying kinetic energy equation;

[tex]E = \frac{1}{2} mv^2\\\\v^2 = \frac{2E}{m} \\\\v= \sqrt{\frac{2E}{m}} \\\\v = \sqrt{\frac{2(4.2 \times 10^{-21})}{7.31\times 10^{-26}}}\\\\v = 338.99 \ m/s \ \approx 339 \ m/s[/tex]

Thus, the root-mean square speed is 339 m/s.

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The molality of a 98.1% sulfuric acid solution is calculated to be 526.32 mol/kg and the molarity is found to be 18.32 M.

To find the molality and molarity of sulfuric acid in a solution, we need to know the mass and volume of the solution. Given that the solution is 98.1% sulfuric acid by mass, we can say that for every 1000g (or 1kg) of solution, there are 981g of sulfuric acid. The molecular weight of sulfuric acid is 98.08 g/mol, so we can convert this to moles to find the molality.

Molality = moles of solute / mass of solvent in kg

The moles of sulfuric acid = 981g / 98.08 g/mol = 10 mol. The mass of the water in the solution is 1000g - 981g = 19g, or 0.019kg. So, the molality of the solution is 10 mol / 0.019 kg = 526.32 mol/kg.

To find the molarity, we need to know the volume of the solution. Given the density of the solution is 1.83 g/ml, we can say that 1kg of solution is 1000g / 1.83g/ml = 546ml = 0.546L. So, the molarity = moles of solute / volume of solution in liters = 10 mol / 0.546 L = 18.32 M.

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The balanced reaction would be written as:
C7H6O3 + C4H6O3--->C9H8O4 + HC2H3O2
To determine the percent yield, we need to first determine the theoretical yield if the reaction were to proceed completely. Then, we divide the actual yield that is given to the theoretical yield times 100 percent. The limiting reactant from the reaction would be salicylic acid. We do as follows: 

Theoretical yield: 1.15 g C7H6O3 ( 1 mol / 138.12 g ) ( 1 mol C9H8O4 / 1 mol C7H6O3 ) ( 180.17 g / mol ) = 1.50 g C9H8O4 should be produced

Percent yield = 1.21 / 1.50 x 100 = 80.66 %

Thus, only 80.66% of the theoretical C9H8O4 is being produced.

The percent yield of the reaction is 14549%.

To find the percent yield, we need to compare the actual yield (1.21 g of acetylsalicylic acid) to the theoretical yield, which is the maximum amount of acetylsalicylic acid that could be produced based on the reactants and their stoichiometry.

We can calculate the theoretical yield using the molar ratios of the reactants. The balanced equation for the reaction is:

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

From the equation, we can see that 1 mole of salicylic acid reacts with 1 mole of acetic anhydride to produce 1 mole of acetylsalicylic acid. Therefore, we can use the moles of salicylic acid and acetic anhydride to calculate the theoretical yield of acetylsalicylic acid.

The moles of salicylic acid can be calculated using its mass and molar mass:

moles of salicylic acid = mass of salicylic acid / molar mass of salicylic acid = 1.15 g / 138.12 g/mol = 0.00833 mol

The moles of acetic anhydride can be calculated using its mass and molar mass:

moles of acetic anhydride = mass of acetic anhydride / molar mass of acetic anhydride = 2.83 g / 102.10 g/mol = 0.0278 mol

Since the balanced equation shows a 1:1 molar ratio between salicylic acid and acetylsalicylic acid, the moles of salicylic acid and acetic anhydride are equal to the moles of acetylsalicylic acid produced.

Therefore, the theoretical yield of acetylsalicylic acid is 0.00833 mol.

The percent yield can be calculated using the formula:

percent yield = (actual yield / theoretical yield) * 100

Substituting the values, we get:

percent yield = (1.21 g / 0.00833 mol) * 100 = 14549%

The percent yield is 14549%.

The molality of solution formed by dissolving 34.8 g of LiI in 500 mL of water is [tex]\boxed{0.520{\text{ m}}}[/tex].

Further Explanation:

Concentration terms are used to determine concentration of various components present in any mixture. Some of these are mentioned below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

7. Parts per billion (ppb)

Molality is defined as moles of solute that are present per kilograms of solvent. It is represented by m and its unit is mol/kg. The expression for molality of solution is as follows:

[tex]{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}[/tex]  

Given information:

Mass of LiI: 34.8 g

Volume of water: 500.0 mL

To calculate:

Molality of LiI solution

How to proceed:

Step1: Moles of LiI that are present in 34.8 g of LiI has to be determined. This is done with the help of equation (1).

The formula to calculate the moles of LiI is as follows:

[tex]{\text{Moles of LiI}} = \dfrac{{{\text{Mass of LiI}}}}{{{\text{Molar mass of LiI}}}}[/tex]                                            …… (1)

The mass of LiI is 34.8 g.

The molar mass of LiI is 133.841 g/mol.

Substitute these values in equation (1).

[tex]\begin{aligned}{\text{Moles of LiI}} &= \frac{{{\text{34}}{\text{.8 g}}}}{{{\text{133}}{\text{.841 g/mol}}}} \\&= 0.26{\text{ mol}} \\\end{aligned}[/tex]  

Step 2: Mass of water is to be evaluated.

At standard temperature and pressure conditions, one liter of water can be considered equivalent to 1 kilograms of water.

Therefore mass of water can be calculated as follows:

[tex]\begin{aligned}{\text{Mass of water}} &= \left( {500.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)\left( {\frac{{1{\text{ kg}}}}{{1{\text{ L}}}}} \right) \\&= 0.5{\text{ kg}} \\\end{aligned}[/tex]  

Step 3: Molality of LiI solution is to be calculated. This is done with the help of equation (2).

The formula to calculate molality of LiI solution is as follows:

[tex]{\text{Molality of LiI solution}} = \dfrac{{{\text{Moles of LiI}}}}{{{\text{Mass of water}}}}[/tex]                                                …… (2)

Substitute 0.26mol for moles of LiI and 0.5 kg for mass of water in equation (2).

[tex]\begin{aligned}{\text{Molality of LiI solution}} &= \frac{{{\text{0}}{\text{.26 mol}}}}{{{\text{0}}{\text{.5 kg}}}} \\&= 0.520{\text{ m}} \\\end{aligned}[/tex]  

Hence, molality of given solution comes out to be 0.520 m.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, solutions, molarity, molality, LiI, moles, mass, molar mass, 0.520 m, 0.5 kg, 0.26 mol.

Final answer:

To distinguish between pure water, salt water, and sugar water, conduct tests such as adding silver nitrate to detect chloride ions or observe evaporation rates in a sealed chamber. Salt water may form a precipitate upon adding silver nitrate, while pure water has a higher vapor pressure than sugar water.

Explanation:

To distinguish between a beaker of pure water, salt water, and sugar water, we need to conduct simple experiments that reveal the properties of each liquid as they differ in terms of their chemical components and physical properties.

Adding silver nitrate to a clear colorless solution that forms a pale yellow precipitate indicates the presence of chloride ions, suggesting a salt water solution. A beaker containing pure water would not react, while sugar water would also not produce a precipitate.

When considering evaporation in a sealed chamber, the beaker with pure water will have a higher vapor pressure compared to the sugar water due to the presence of sugar molecules at the surface, which inhibit the evaporation rate. Over time, all of the water from the beaker with higher vapor pressure (pure water) will evaporate and condense in the beaker with the sugar solution.

To differentiate between the three beakers, we must observe the reactions mentioned above (if testing chemically) or deduce based on the properties like vapor pressure and the effects they have on evaporation and condensation processes.

Given:

Mass of H2 = 5.22 kg = 5220 g

Mass of N2 = 31.5 kg = 31500 g

To determine:

Theoretical yield of NH3

Explanation:

The balanced chemical reaction is:

N2 + 3H2 → 2NH3

1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3

# moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles

# moles of H2 = 5200 g/ 1 g.mol-1 = 5200 moles

Therefore N2 is the limiting reagent

Based on the stoichiometry:

1 mole of N2 forms 2 moles of NH3

Thus, 1125 moles of N2 will yield : 1125 * 2 = 2250 moles of NH3

Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg

Ans: Theoretical yield of NH3 = 38.3 kg

Answer : The  theoretical yield of ammonia is, 29.58 Kg

Solution : Given,

Mass of [tex]H_2[/tex] = 5.22 Kg  = 5220 g

Mass of [tex]N_2[/tex]= 31.5 Kg = 31500 g

Molar mass of [tex]H_2[/tex] = 2 g/mole

Molar mass of [tex]N_2[/tex] = 28 g/mole

Molar  mass of [tex]NH_3[/tex] = 17 g/mole

First we have to calculate the moles of [tex]H_2[/tex] and [tex]N_2[/tex] .

Moles of [tex]H_2[/tex]= [tex]\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{5220g}{2g/mole}=2610moles[/tex]

Moles of [tex]N_2[/tex] = [tex]\frac{\text{ given mass of }N_2}{\text{ molar mass of }N_2}= \frac{31500g}{28g/mole}=1125moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

From the balanced reaction we conclude that

1 mole of [tex]N_2[/tex] react with 3 moles of [tex]H_2[/tex]

1125 moles of [tex]N_2[/tex] react with [tex]3\times 1125=3375[/tex] moles of [tex]H_2[/tex]

That means [tex]H_2[/tex] is a limiting reagent and [tex]N_2[/tex] is an excess reagent.

Now we have to calculate the moles of ammonia.

From the reaction we conclude that,

3 moles of [tex]H_2[/tex] react to give 2 moles of ammonia

2610 moles of [tex]H_2[/tex] react to give [tex]\frac{2}{3}\times 2610=1740[/tex] moles of ammonia

Now we have to calculate the mass of ammonia.

[tex]\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3[/tex]

[tex]\text{Mass of }NH_3=(1740moles)\times (17g/mole)=29580g=\frac{29580}{1000}=29.58Kg[/tex]

Therefore, the  theoretical yield of ammonia is, 29.58 Kg

In the formation of ammonia, different initial amounts of reactants result in different Kc values at equilibrium. While the temperature is constant, varying concentrations of reactants and products lead to different equilibrium constants Kc.

The chemical equation is given by 2NH3(g) = N2(g) + 3H2(g). In the first scenario, you start with 1.30 mol of N2 and 1.65 mol of H2 in a 1.00 L container, with 0.100 mol of NH3 at equilibrium. This implies a decrease in N2 and H2 by x amount and an increase in NH3 by 2x. Given the stoichiometric relationships, we can calculate that the equilibrium concentrations of N2 and H2 are 1.20 M and 1.30 M respectively.

Subsequently, we can calculate Kc using the equilibrium concentrations, Kc = [NH3]^2/(N2][H2]^3) = (0.10)^2/(1.20 * (1.55)^3) = 422 M^-2.

In the second scenario, the same calculations result in a Kc = 20.5 M^-2. The discrepancy in the Kc values arises due to the differences in the initial amounts of reactants and the products formed. Although the temperature is constant and the reaction is the same, the amount of reactants used and the products formed vary which impacts the Kc value.

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0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

Final answer:

To obtain a 100 mL solution of 0.300 M KI, it would take approximately 4.98 grams of KI. To create a 100 mL solution of 0.600 M KI, it would take approximately 9.96 grams of KI.

Explanation:

To determine how many grams of KI are needed to create a 100 mL solution of 0.300 M KI, we can use the equation:

Molarity (M) = moles of solute / volume of solution (L)

First, convert the volume from mL to L: 100 mL = 0.100 L

Then, rearrange the equation to solve for moles of solute:

moles of solute = Molarity x volume of solution

moles of solute = 0.300 M x 0.100 L

moles of solute = 0.030 mol KI

Finally, use the molar mass of KI (166 g/mol) to convert moles to grams:

grams of solute = moles of solute x molar mass

grams of solute = 0.030 mol KI x 166 g/mol

grams of solute = 4.98 g KI

Therefore, it would take approximately 4.98 grams of KI to obtain a 100 mL solution of 0.300 M KI.

Similarly, to create a 100 mL solution of 0.600 M KI, the calculation would be:

moles of solute = 0.600 M x 0.100 L

moles of solute = 0.060 mol KI

grams of solute = 0.060 mol KI x 166 g/mol

grams of solute = 9.96 g KI

It would take approximately 9.96 grams of KI to create a 100 mL solution of 0.600 M KI.

The base dissociation constant or Kb is a value used to measure the strength of a specific base in solution. To determine the percent ionization of the substance we make use of the Kb given. Methylamine or CH3NH2 when in solution would form ions:

CH3NH2 + H2O < = >  CH3NH3+ + OH- 

Kb is expressed as follows:

Kb = [OH-] [CH3NH3+] / [CH3NH2]

Where the terms represents the concentrations of the acid and the ions. 

By the ICE table, we can calculate the equilibrium concentrations,
     CH3NH2    CH3NH3+             OH- 
I      1.60           0                         0
C      -x               +x                      +x
 --------------------------------------------------
E  1.60-x             x                       x

Kb = [OH-] [CH3NH3+] / [CH3NH2] =  3.4×10−4

Solving for x,

x = [OH-] = 0.023 M 

pH = 14 + log 0.023 = 12.36

Therefore, the first option is the closest one.

Answer:

5.2

Explanation:

Calculate the pH of a 1.60 M CH₃NH₃Cl solution. Kb for methylamine, CH₃NH₂, is 3.7 × 10⁻⁴.

CH₃NH₃Cl is a strong electrolyte that ionizes according to the following equation.

CH₃NH₃Cl(aq) → CH₃NH₃⁺(aq) + Cl⁻(aq)

The concentration of CH₃NH₃⁺ will also be 1.60 M (Ca). CH₃NH₃⁺ is the conjugate acid of CH₃NH₂. We can find its acid dissociation constant (Ka) using the following expression.

Ka × Kb = Kw

Ka × 3.7 × 10⁻⁴ = 1.0 × 10⁻¹⁴

Ka = 2.7 × 10⁻¹¹

The acid dissociation of CH₃NH₃⁺ can be represented through the following equation.

CH₃NH₃⁺(aq) ⇄ CH₃NH₂(aq) + H⁺(aq)

For a weak acid, we can find the concentration of H⁺ using the following expression.

[H⁺] = √(Ka × Ca) = √(2.7 × 10⁻¹¹ × 1.60) = 6.6 × 10⁻⁶ M

The pH is:

pH = -log [H⁺] = -log (6.6 × 10⁻⁶) = 5.2

There are approximately 2.08 x 10^25 hydrogen atoms in 5.20 mol of ammonium sulfide.

To determine the number of hydrogen atoms in 5.20 mol of ammonium sulfide (NH4)2S, we need to consider the molar ratio of hydrogen atoms to moles of ammonium sulfide. In the formula NH4)2S, there are 8 hydrogen atoms per 1 molecule. To find the number of hydrogen atoms, multiply the number of moles of ammonium sulfide by the Avogadro's number (6.022 x 1023 atoms/mol) and then multiply by the number of hydrogen atoms per molecule. So, the number of hydrogen atoms in 5.20 mol of ammonium sulfide is:
5.20 mol x 6.022 x 1023 atoms/mol x 8 atoms = 2.08 x 1025 hydrogen atoms.

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In [tex]5.20[/tex] moles of ammonium sulfide, there are approximately  [tex]2.51 x 10^{25[/tex] hydrogen atoms.

To determine the number of hydrogen atoms in [tex]5.20[/tex] moles of ammonium sulfide [tex](\text{NH}_4)_2\text{S}[/tex], we must comprehend the compound's makeup.

This indicates it contains two ammonium ions [tex]\text{(NH}_4^+)[/tex] and one sulfide ion ([tex]\text{S}^{2-}[/tex]).

Since there are two ammonium ions in each formula unit of ammonium sulfide, there are a total of 8 hydrogen atoms per formula unit of [tex](\text{NH}_4)_2\text{S}[/tex].

Each mole of[tex](\text{NH}_4)_2\text{S}[/tex] contains Avogadro's number ([tex]6.022 \times 10^{23[/tex]) of formula units.

Thus, [tex]5.20[/tex] moles of [tex](\text{NH}_4)_2\text{S}[/tex] contain:
[tex](5.20 \, \text{moles}) \times (6.022 \times 10^{23} \, \text{formula units/mole}) = 3.13144 \times 10^{24} \, \text{formula units}[/tex]

The concentration of the KCl solution is 1.61 molal.

The concentration of a solution can be described by its molality (m), which is the number of moles of solute per kilogram of solvent. To find the molality of the KCl solution, we need to calculate the number of moles of KCl and the amount of water in kilograms. Given that 1.43 mol of KCl is dissolved in 889 g of water, we convert the mass of water to kilograms (889 g = 0.889 kg). Now we can calculate the molality:

Molality (m) = moles of KCl / kilograms of water

Molality (m) = 1.43 mol / 0.889 kg = 1.61 molal

the atomic weight of the element : 122.22 amu

The elements in nature have several types of isotopes

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom

Mass atom X = mass isotope 1 . % + mass isotope 2.%

An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.

The 'amu' unit has now been replaced with a unit of 'u' only

for example, Carbon has 3 isotopes, namely ₆¹²C, ₆¹³C, and ₆¹⁴C

Element has two isotopes. 120.9038 amu with 57.25% abundance and the other has an atomic weight of 122.8831 amu

Mass atom element X = mass isotope 1 . % + mass isotope 2.%

Mass atom element X = 120.9038 . 57.25% + 122.8831. 42.75%

Mass atom element X = 69.22 + 52.53 = 122.22 amu

The subatomic particle that has the least mass

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Keywords: mass number, atomic mass, amu, isotope

There are 3 atoms in total. Two come from Hydrogen hence the H2. The third comes from the single Oxygen (O). Hope this helps!

Three atoms make up the water molecule (H2O) as a whole.

According to the context, the term may or may not include ions that meet this requirement. A molecule is a collection of two or more atoms held together by attractive forces known as chemical bonds.

The same atoms can combine in various ratios to create various molecules. For instance, water (H2O) is made up of two hydrogen atoms and one oxygen atom, whereas hydrogen peroxide (H2O2) is made up of two hydrogen atoms and two oxygen atoms.

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Answer: determining the concentration of an unknown base

Explanation:  The last step in performing a titration will lead to the determination of the concentration of the unknown base or an unknown quantity.

The other steps like finding out  which pH indicator works,  finding the number of moles of a product produced in a reaction and determining the molecular masses of the products in the reaction comes in between the procedure.

Answer: The balanced chemical equation is written above.

Explanation:

Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.

When ferrous sulfate reacts with lead (II) nitrate, leads to the production of iron (II) nitrate and lead (II) sulfate.

The chemical equation for the above reaction follows:

[tex]FeSO_4(aq.)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+PbSO_4(s)[/tex]

By Stoichiometry of the reaction:

1 mole of aqueous solution of ferrous sulfate reacts with 1 mole of aqueous solution of lead (II) nitrate to produce 1 mole of aqueous solution of iron (II) nitrate and 1 mole of solid lead (II) sulfate.

Hence, the balanced chemical equation is written above.

The root mean square velocity (RMS velocity) is a method to find a single velocity value for the particles. The average velocity of gas particles in a mixture is found using the root mean square velocity formula:

 μrms = sqrt (3RT/M)

Where,

R = universal gas constant = 8.3145 (kg·m^2/s^2)/K·mol

T = temperature = 35.9 ˚C = 309.05 K

M = molar mass = 28 * 10^-3 kg / mol

Substituting the given values into the equation:

μrms = sqrt [(3 * 8.3145 (kg·m^2/s^2) /K·mol) * (309.05 K) / (28 * 10^-3 kg / mol))]

μrms = sqrt (275,313.8813 m^2)

μrms = 524.7 m / s                     (ANSWER)

The appropriate formula for solving this problem includes the variables pressure and temperature (P, T).

To solve this problem, we can use the combined gas law formula, which includes the variables pressure (P), volume (V), and temperature (T).

The combined gas law formula is as follows: P1V1/T1 = P2V2/T2

In this problem, the volume is constant, so we only need to consider pressure and temperature. Therefore, the appropriate formula for solving this problem includes the variables pressure and temperature (P, T).

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1. After calculation, the new pressure of the gas is approximately 1.139 atm. 2.The variables involved are pressure (P) and temperature (T).

1. To determine the new pressure of the gas after the temperature increase while keeping the volume constant, we should use the combined gas law, which states:

P₁/T₁ = P₂/T₂

Given:

Rearrange the formula to solve for P₂:

[tex]\[P_2 = P_1 \cdot \frac{T_2}{T_1}\][/tex]

Substitute the given values:

[tex]P_2 = 1 \, \text{atm} \cdot \left( \frac{311 \, \text{K}}{273 \, \text{K}} \right)[/tex]

Calculate:

P₂ ≈ 1.139 atm

Therefore, the new pressure will be approximately 1.139 atm.

2. The most appropriate formula for solving this problem includes only the following variables: P, T.

Mole measure the number of elementary entities of a given substance that are present in a given sample. The mass of 4.00 mole of sodium chloride is 233.76g.

The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity of amount of substance. One mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.

Mathematically,

mole of sodium chloride = mass ÷ Molar mass of sodium chloride

number of mole of sodium chloride=4mol

Molar mass sodium chloride =58.44 g/mole

Substituting the given values in the above equation, we get

4mole of sodium chloride=mass mass of sodium chloride÷ 58.44 g/mole

mass of sodium chloride = 4× 58.44 g/mole

mass of sodium chloride= 233.76g

Therefore the mass of 4.00 mole of sodium chloride is 233.76g.

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528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.

Heat is the result of the movement of kinetic energy within a material or an item, or from an energy source to a material or an object. Radiation, conduction, and convection are the three mechanisms through which such energy can be transferred.

Heat is the quantity of energy that is transferred from one structure to its surrounds as a result of a change in temperature. Heat is the transfer of kinetic energy between one media or item and another, or from an energy source to a medium or object.

The total amount of heat required

∑q = q1 + q2 + q3

q1 = mCΔT

= 475 g x 0.902 J/g-oC x ( 933.47 - 295 ) K

= 273,552.47J

= 273 kJ    

ΔT =  ( 933.47 - 295 ) K

= 638.47 K  

ΔT = ( 933.47 K -273) - (295 K - 273)

= 638.47°C

For ΔT temperature in K is same as in °C

q2 = mol ΔHf  

mol = 475 g x 1 mol /27g

= 17.6 mol    

q2 = mol ΔHf = 17.6 mol x 10.79 kJ/mol

= 189.9 kJ

q3 = mCΔT

= 0.475 kg x 1.18 kJ/kg-oC (1050 - 933.47)K

= 65.3 kJ

∑q = q1 + q2 + q3

= 273 + 189.9 + 65.3

= 528.2 kJ

Thus,528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.

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Final answer:

To calculate the molality of the solution, divide the number of moles of lithium chloride (0.990 moles) by the mass of the water in kilograms (3.6 kg), resulting in a molality of 0.275 m.

Explanation:

The molality of a solution is calculated by finding the number of moles of solute and dividing by the mass of the solvent in kilograms. To find the molality of the solution with 42 grams of lithium chloride (LiCl) dissolved in 3.6 kg of water, we first convert grams to moles using the molar mass of LiCl.

The molar mass of LiCl is 42.39 g/mol. So, if we divide 42 grams by the molar mass, we get the number of moles of LiCl:

Moles of LiCl = 42 g / 42.39 g/mol = 0.990 moles of LiCl

We then use this number of moles and the mass of the solvent in kilograms to find the molality:

Molality (m) = Moles of solute / Mass of solvent (kg) = 0.990 moles / 3.6 kg = 0.275 m

Therefore, the molality of the solution is 0.275 m.

equation is

pH=14+log(.2)